This invention relates to superconducting machines, and more particularly to such a machine having a cooling system and to a method of operating such a machine.
Superconducting electrical machines require a cryogenic cooling system to keep at least one of their major components (i.e. the rotor or the stator) at cryogenic temperatures suited to the superconducting state. Thus the cryogenic system allows at least one of the electrical machine's windings (i.e. on the stator and/or on the rotor) to operate in the superconducting state. The cryogenic system is usually sized to remove the steady-state heat load (i.e. the heat leaking into the cryogenically cooled region from warmer regions) from the superconducting major component, usually the rotor.
Superconducting electrical machines of large power and low speed (e.g. of the order of a few megawatts running at up to a few hundred r.p.m.) such as are used in marine propulsion are physically large and have large thermal mass. In consequence the superconducting machine takes a long time to cool down from ambient to the temperature required for the superconducting state. This cool-down period is of the order of days in machines rated as above.
Compared to a conventional (i.e. non-superconducting) electrical machine, the long cool-down period represents a commercial disadvantage which works against the widespread adoption of large superconducting machines by the market. This arises because the superconducting electrical machine cannot be used until its relevant major component has been cooled to the required cryogenic temperature, but it is uneconomic to keep it cooled permanently.
Simple analysis shows that the time to cool down a body of mass M, initially at temperature Thot and subsequently at temperature T after time t, is given by equation 1:—t(T)=((M·S)/(h·A))·ln [(Thot−Tcool)/(T−Tcool)]  Equation 1.
where ‘ln’ is the logarithm to the base ‘e’, S is the specific heat of the body, A is the area of the body that is exposed to the coolant, h is the heat transfer coefficient (assumed constant) from the body to the coolant, and Tcool is the temperature (assumed constant) of the coolant. This formula shows that the cool-down period (t) is reduced if the surface area of the body exposed to the coolant is increased and/or the heat transfer coefficient is increased. Similarly, the cool-down period is reduced if the temperature of the coolant (Tcool) is reduced.
The above formula can be recast in dimensionless form as:—h·A·t(T)/(M·S)=ln [[(Thot−Tcool)/(T−Tcool)]  Equation 2.
If initial temperature Thot=300K (i.e. approx. room temperature), Tcool=20K and the target temperature T=25K. Then the equation gives ln(280/5)=4.03 approximately.
If the coolant temperature is reduced so that Tcool=4.2K (i.e. liquid-helium temperature) then ln(295.8/20.8)=2.66 approx. which represents a (time) saving of about one third (1−2.66/4.03). However, a colder coolant is obviously more expensive.
In general terms, for maximum cooling it is desirable (i) to increase the body area (A) which is exposed to the coolant, (ii) to increase the heat transfer coefficient (h) from the body to the coolant, and (iii) to reduce the coolant temperature (Tcool), but all these measures are likely to involve increased expense.